Download Narrowing down the candidate of the NAE (nuclear active

Narrowing down the candidate of the
NAE (nuclear active environment) of
the nuclear cold fusion.
Tetsuo Sawada (Nihon U.)
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Three important properties of the nuclear CF
(1) Openning of the channel d
 d  4He
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at zero incident energy.
(2) Closing of the channels d  d  t  p and d  d  He  n
,
which are the main channels of the low energy d+d reaction in vacuum.
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(3) Drastic lowering of the repulsive Coulpmb potential between d's,
which prevents for the two deuterons to come close and to fuse.
We shall see that the first property is sufficient to narrow down the
candidate of the NAE of the nuclear cold fusion uniquely, and so the
second and third properties will be derived from the first one.
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energy-momentum conservation in vacuum
• In the d + d → t + p reaction, energy Q is
shared by the final state particles of
mass m1 and m2 as
2
p'
2
 2
Q  p' / 2 m 1 ( p' ) / 2 m 2
in the centr of mass system. Namely
triton gets 25% whereas p gets 75% of Q.
• However when the mass is zero, we
must use the relativistic form of energy
E  p 2 c 2  m 2 c 4 instead. So the
energy equation of d+d →α +γ is
Q  p' c  p'2 / 2M 4 , whose solution is
p' c / Q  1  (1  1  2Q / M 4 ) /(1  1  2Q / M 4 )
 0.99682
- p'
Q = 23.9 MeV.
( Q-value)
M4 = 3732 MeV. ;
(mass of alpha particle)
therefore only 0.318% of Q becomes the
kinetic energy of the alpha particle.
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energy-momentum conservation of nuclear CF
Let us examine the conservation law of
d+d → He(4) of the nuclear CF, or generally
(two body) → (one body) type. It will be
shown that the consevations of energy and
momentum are not compatible.
p'
p
proof: in the center of mass system, the
momentum p' of the final state particle is zero
from the momentum conservation, whereas
from the energy conservation the magnitude
of p' must satisfy | p'| ²=(2MQ).
qed.
Therefore the reaction of the nuclear CF
indicates the existence of the external
potential outside of the nuclear system,
which receives the momentum transfer.
-p
p'
- p'
external potential
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Narrowing down the candidates of the external potential 1
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For our d + d → He(4) reaction, the momentum transfer q is as large as
q=422 MeV./c , whose De Broglie wave length is λ/2π = ħ/q =0.56fm.. So
we expect that the external potential cannot be soft and spread but it
must be sharply localized in order to do the job to receive such large q.
More precisely, the probability of the momentum transfer q by the external
potential Vext( r) is the square of the scattering amplitude f(q) given by :
 
f (q)  (2m / 4 )  d 3r exp[ iq  r ] Vext (r )
which is essentially the Fourier transformation of the external potential.
When the variations of the exponential factor and the potential Vext coincide,
f(q) become reasonablly large. Next example will help to understand the
situations.
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Narrowing down the candidates of the external potential 2
When the electric charge distribution is exponential form of
 (r )  C exp[  br ]
 ( R) 
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the external potential has the form:
1
bR 
bR
1

e
(
1

)

R
2 
which reduces to the Coulomb potential at large R.
The Fourier tranformation of the potential φ(R) is :
~ (q) 
1
1
q 2 (1  q 2 / b 2 ) 2
in which b and √< r ²> relates by < r ²>=12/b².
Since the size of the wawe packet of the electron cloud is the order of
1Å, and ħ/q is 0.56 fm. in our case, q/b ≈ 60000. Therefore | f(q)|²
becomes extreamly small. So the electron cloud cannot do the job to
absorb such a large momentum transfer q of the d+d → He(4) reaction.
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unique candidate of the external field 1
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There are two candidates of the external field, to which the nucleus can
responds. They are the electric field E and the magnetic field B, because
the nucleus has two attributes, namely the (positive) electric charge Ze
and the magnetic dipole moment μ. The interaction energy V of these
attributes and the external fields are Zeφ and - μ•B, respectively, where
φ is the scalar potential of the electric field: E= - gradφ . Among them,
we already observed that E produced by the electron cannot do the job.
On the other hand, since the magnetic Coulomb field produced by the
magnetically charged particle (magnetic monopole) is B=*e/r² , the
external potential of the nucleon Vm is:

Vm (r )   tot (e / 2m p )(*e / r )   rˆ F (r ) where
2
F (r )  1  e  ar (1  ar  a 2 r 2 / 2) with a  6.04 .
, in which F(r) is the form factor of the nucleon.
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unique candidate of the external field 2
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The Fourier transformation of the external potential Vm(r) is
~
Vm (q)  i  tot (*ee / 2m p )( 2 / a)
(a / q) 5

  qˆ
2
2 2
(1  a / q )
However since, in our d+d → He(4) reaction q = 422MeV./c =3.02μπ
and a=6.04μπ , the ratio becomes (a/q)=2.0. Therefore the Fourier
transfomation of Vm(r) stays resonablly large.
Contrarily to the case of the electric field produced by the electron
cloud, the momentum transfer q from the nuclear system goes
smoothly when the external field is the magnetic field produced by
the magnetic monopole.
In this way the first mystery of the nuclear CF is solved if the fusion
reaction proceeds under the influence of the magnetic monopole.
Therefore NAE is the magnetic field produced by magnetic monopole.
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Summary ( How the magnetic monopole is selected as NAE )
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We repeat:
(1) In the nuclear CF, the channel of d + d → He(4) reaction opens.
However this reaction violates the conservation law of the momentum
of the nuclear system. Momentum q must be transfered to the external
field. Numerically |q| is as large as 422.0 MeV./c in our case.
(2) Since the nucleus must responds to the external field, we can
narrow down the candidates of the fields further. They are the electric
field or the magnetic field, which respond to the electric charge Ze and
the magnetic dipole moment μ of the nucleus respectively.
(3) Because the magnitude of q is so large, |q|=422MeV./c , whose
deBroglie wave length is 0.56fm., the external field must be sharply
localized. The electric field produced by the packet of electron , whose
size is limitted by the Compton wave length of electron (386fm.)
cannot do the job.
(4) There remains only the magnetic field B produced by the
magnetic monopole. We shall see that all other mysteries are
resolved by the external magnetic field prduced by *e.
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Charge quantization condition of Dirac and the superstrong nature of the magnetic Coulomb field
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If we remember that the angular momentum density of the electro-magnetic
field is r×(E×H)/c , the total angular momentum of the system, where the
electric charge Q and the magnetic charge *Q coexist, stored in space is
M = - (*QQ/c) e, in which e is the unit vector connecting two charges. Since
in the quantum theory, a component of the angular momentum must be the
integer multiple of ℏ/2 , we arrive at the "charge quantization condition" of
Dirac : (*QQ/ℏc)=n/2 with n=0, ±1, ±2, ⋯.
If we substitute the charges *Q and Q by the smallest non-zero charges *e
and e, then we get the relation between the fine structure constant and its
magnetic counterpart : (*e²/ℏc)(e²/ℏc)= 1/2. Since (e²/ℏc)= 1/137.04 , we
obtain (*e²/ℏc)= 137.04/4, wich is super-strong. Therefore the external potential
Vext becomes strong enough to receive the momentum transfer with high probability.
Another important role of the super-strong monopole is that the deuterons
are trapped by the monopole deeply enough to be prevented to transit to
the continuous t+p or He(3)+n channels. We shall see this in the next page.
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How the main channels d+d → t+p, d+d →He(3)+n are closed ----- derivation of the property (2) of the nuclear CF. ------If we remember the binding energies of d, t and He(3) are 2.2, 8.5 and
7.7 MeV. respectively, the Q-values of d+d → t+p and d+d → He(3)+n are
4.1 and 3.3 MeV. respectively.
When the nuclear CF proceeds after the two deuterons are trapped by
the magnetic monopole *e, the energies of the final states decrease as
the binding energies of the *e-d namely B*e-d increase. If B*e-d exceeds
Q/2, the energy of the final state becomes negative. Therefore the
channel is closed.
To find the binding energy of the ground state of the *e-d system, the
variational calculation is carried out with the Hamiltonian:
H  H  H  VN
p
0
n
0
 2

1

with H 0 
(i  ZeA)   tot (*ee / 2m p )(  rˆ) / r 2
2m
, where A is the vector potential and Z=1 for proton and Z=0 for
nuetron. The result is B=2.3 MeV.. Therefore channels are closed.
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Energy levels of the A=4 and Z=2 channels
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Since the binding energy of thedeuteron is 2.2 MeV., the threshold of
of the d+d channel is E= - 4.4 MeV. and the sky blue belt of the next
figure indicates the continuous spectrum of the d+d channel.
Likewise, E= - 8.5 MeV. and E= -7.7 MeV. are the threshold of the t+p
and He(3)+n channel respectively. Energy of the state where all the
nucleons are separated infinitely is chosen as the origin of E (E=0).
In vaccum, if we start the d+d reaction with the zero kinetic energy, its
energy level is E= - 4.4 MeV.. From the energy conservation, this
energy level means the kinetic energy of the t+p and He(3)+n channel
is 4.1 and 3.3 MeV. respectively, and so such reactions can occur. On
the other hand, if the d+d reaction starts after d's are trapped by the
magnetic monopole *e, and form the bound state d-*e-d, then E= -9.0
MeV.. In such a case, kinetic energy of t+p and He(3)+n becomes
negative, which means such reactions cannot occur. However since
He(4) is the tightly bound nucleus, kinetic energy for E= - 9.0 MeV. is
19.2 MeV.. So this channel stay open as long as it satisfies the
momentum conservation.
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d+d
t+p
He(3) + n
He(4)
12
E
MeV.
-4.4
-5
-8.5
-10
-7.7
-7.95
-9.0
-10.02
D=1 monopole
He(4) dominance
-28.2
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Penetration of the repulsive Coulomb barrier 1
The nuclear CF proceeds when the two deuterons are trapped by the
same magnetic monopole *e, whose orbital sizes are several fm.. The
trapping of the first deuteron goes rather smoothly. However the
potential V1(x) felt by the second deuteron is the sum of the repulsive
Coulomb and the attractive potential between the magnetic monopole
and the magnetic moment of the nucleus when the spin orients properly.
The explicit form of V1(x) is
V1 ( x)  e2 / x  ( tot / 4mp ) / x 2 ,
in which the charge quantization condition of Dirac, *ee=1/2 , is used.
Because of the attractive second term, the peak value of V1(x) reduces
to 17keV. , which should be compared with the peak value 1MeV. of the
pure Coulomb potential VC(x). However this lowering is not sufficient
to increase the penetration factor to the reasonable value.
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Penetration of the repulsive Coulomb barrier 2
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Further help to increase the penetration factor P comes, when we consider
the electron wave function in the magnetic Coulomb potential. Since the
binding energy is order of the electron mass me, the relativistic treatment
is necessary, so we must solve the Dirac equation in the external magnetic
Coulomb field produced by the magnetic monopole. However I skip this.
If you are interested in this topic, see " Underlying mechanism of the nuclear CF
implied by the energy-momentum consevation" (available at [email protected])
I shall use only the result. The repulsive Coulomb term is shieled to become
V2 ( x)  (e2 / x) exp[ 2me x]  ( tot / 4mp ) / x 2
If we substitute V2(x) to the formula of the penetration factor P for E=0 :
P  exp[ 2 ] with   2red 
b
a
V ( x) dx / 
We obtain P=6.2×10^{-9} , which should be compared with P=5×10^{ -106}
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of the case of VC(x) . Thus the third mystery (3) of CF is resolved.
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One particle catalyst and the sporadic nature of nuclear CF
We saw that the d+d → He(4) reaction occurs when two deuterons form a
bound state with the same magnetic monopole *e. Since such a state is
unstable and transits to more stable α-particle. Because the spin of α is
zero, it is not attracted by *e. The magnetic monopole simply emits the
alpha particle, and there remains fresh monopole *e, and which starts to
attracts surrounding fuel deuterons again. In this way the magnetic
monopole plays the roll of the " one-particle catalyst " of the nuclear CF.
At present, since we rely on the floating magnetic monopole in nature and
it is a very rare particle, we must wait for long time before the monopole
*e comes into and stops in the region where the deuteron density is
sufficiently high. Therefore the starting of the nuclear CF is governed by
the probability.
We may expect to facilitate the nuclear CF, if we can succeed to gather
the floating magnetic monopole to the region where the deuteron
density is high. In the next we shall consider this possibility.
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Proposal of experiments and devices 16
for the magnetic monopole
• Devices for magnetic monopole:
(A) trapper : necessary to store the magnetic monopole.
(B) detector :
super-conducting ring can do the job.
(C) collector : magnetic dipole can collect the floating
magntic monopole.
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(A) Magnetic monopole *e trapped in the lattice 1
In order to learn the process of trapping
of the magnetic monopole *e by a lattice,
by a rare earth for example, let us
compute the trapping energy ΔE by using
a simple model.
We consider a lattice
of atoms with large magnetic moment
κ(e/2me) whose directions are random
(see upper figure). When a magnetic
monopole *eD comes in, the magnetic
moments align spherically (down figure).
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Before *e comes in.
Because of the thermal agitation, the
directions of the spins are random
outside of the blue sphare of radius R,
and R is determined by:
e * eD
1


k BT
2
2me R
2
where kB is the Boltzmann constant.
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After *e comes in and stops.
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(A) Magnetic monopole *e trapped in the lattice 2
The trapping energy ΔE is the sum of the contributions of the magnetic
moment κ mesured by the unit of Bohr magneton e/2me in the
magnetic Coulomb field *eD/r ² . D is 1 and 2 for Dirac and for
Schwinger magnetic monopole respectively.
4
E  3
a
R
r
0
2
dr
 e *e D
2me
r
2

D R
a
3
me
where a is the lattice constant.
When the lattice a is 1Å , the numerical values of ΔE and R are:
D
3/2 14.24 keV
E  (
)
,
2
T
R
 D 2.97 10  6 cm
2
T
It is remarkable that ΔE and R are proportinal to 1/√T. Since at room
temperature T=300˚, ΔE is the order of 1keV., and R is few nm..
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(B) Detector of the magnetic monopole 1
Total magnetic flux Δϕ produced
by the magnetic charge *eD is
Δϕ= 4π*eD. Therefore when
the magnetic charge *eD passes
through the super-conducting ring
of the torus shape as indicated by
the right figure, the magnetic flux
trapped by the ring must increase
by Δϕ.
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*e D
Since the the super-conducting
current I circulating the inner surface
of the ring is proportional to the total
magnetic flux ϕ trapped by the ring,
we can determine the change of the
flux Δϕ by observing the change of
the surface current ΔI.
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(B) Detector of the magnetic monopole 2
We change the equation


rot H  4 i / c
to the integral form, in which the red
line of the figure is the contuor C of
Stokes theorem. Because of Meisner
effect that the magnetic field inside of
the super-conductor is zero, we get
the relation between the surface
current I and the magnetic field
strength ϕ/S where S is the area of the
inner radius of the torus. Therefore
the change of the current ΔI becomes:
c 2 0 D
D
I 
 0.328
 A / cm
2
4 S
S (cm )
, where ϕ0 is the quantum unit of
the magnetic flux : ϕ0=πℏc/e
and numerically it is
0  2.07 1011 Wb (in MKS ) or
0  2.07 107 G cm2 (in CGS)
ΔΦ
I
C
h
2b
By using the SQUID, the change
of the current can be observed.
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(C) Collector of the of the magnetic monopole *e 1
The magnetic dipole or the solenoid of
the electric current can serve as the
collector of the magnetic monopole,
because the magnetically charged
particle *e with very low velocity moves
on the line of force of the magnetic field.
After Gilbert, we may regared the earth as
a huge magnetic dipole, then we can
expect to find the magnetic monopole with
better probability in the neighborhood of
the magnetic north (or south) pole of earth.
In the next page, we shall consider how to
slow down the speed of the monopole more
to make it be trapped by the potential of
the order of 1 keV. in depth.
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trapper
N
S
22
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Trapping of the magnetic monopole *e 1
Charged particle looses its energy in the
form of radiation, when it enters into the
region of the potential. -dE/dt=(2/3)e²a²
is Larmor's formula of the radiation loss,
in which a is the acceleration. For the
case of the magnetic monople, it must
become:
dE
2

  (e * D ) 2 | a |2
dt
3
Therefore energy loss occurs at the surface
of the metals of high permeability, and so in
order to slow down *e particle effectively, use
of the powder of nm size is desirable. In the
figure the green points are the places where
the radiation is emitted.
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Energy transfer to lattice and hopping of the magnetic monopole
The alpha particle emitted from the magnetic monopole comes to stop
with track length 0.1 mm. The mechanism of the stopping power is
exactly the same as that of the Bethe-Bloch's formula given below.
Namely the energy is transfed to electron first and then to the lattice.
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2me c 2  2 2Tmax
dE
1 1

2 Z
2

 Kz
log
  
2 
dx
A  2
I
2
with K  4 N A e 4 /( me c 2 )  0.307 MeV . g 1 cm 2
Contrarily to the stopping power of the case of the parallel beam, in our
case the energy deposition is not uniform, but heat is stored in the small
neighborhood of the magnetic monopole *e and it becomes the hot spot.
As the temperature raises up, the trapping energy of *e by the lattice
decreases as 1/√T, and the magnetic monopole must hop to cooler
region. Therefore we must observe the discrete hot spots in the
cathode. In fact such discrete hot spots spreading with time were
observed with the infra-red camera by Boss et. al..
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